text:
Can an aircraft owner legally replace an oil quick drain valve?
The quick drain valve on our older Mooney has started leaking and making a bit of a mess.  We went to our local shop to asked them to schedule to replace it for us and they told us they are booked out several weeks and turning away any smaller jobs due to being understaffed -_-
I mentioned we do our own oil changes and he suggested that if we were handy enough to handle an oil change we could just remove the quick drain and replace it with an original plug and he was happy to give us one for free.  That would get us to our next oil change and we could get it replaced then.
Part 43 allows owners to do the oil changes and replacing the quick drain valve with a plug looks to be basically identical to the oil change process if you didn't have a quick drain valve in the first place: remove the plug, catch the oil, replace plug, add new oil.  But if I'm going to down that route anyway why not just grab a new quick drain off of aircraft spruce and put that in instead of the plug?
Are we on the right side of owner maintenance if we replace the quick drain with a plug as the A&P suggested?  How about if we replace the quick drain with a new one?  Or should we get a sign off from an A&P

This is why we need more mechanics on SE.

@RyanMortensen There is a share button that you can use to email this question to somebody.  If you know a good mechanic, you might throw this over to them. Maybe it would entice them to join.

As an A&P familiar with 14 CFR 43, Appendix A, I do not consider replacing an oil quick drain plug to be included in the list of the preventative maintenance that an owner or operator is authorized to do. This is my informed opinion; I am not the FAA.

@JonathanWalters I think you're probably overall right.  Now if I could just solve the "A&P's won't do the small tasks in my area" problem that would be awesome.  We've been outright turned away or told it would be 2-4 weeks for every small issues thing we've wanted to have a mechanic since purchasing.  Seems like the local shops don't want to do anything except annuals and avionics installations that bring the $$$

Another way to ask this is: "how can the FAA determine that someone other than an A&P replaced an oil quick drain valve?"

This is what the FAR allows you to do as preventive maintenance if you are not a certified mechanic:

(c) Preventive maintenance. Preventive
maintenance is limited to the following
work, provided it does not involve complex
assembly operations:
(1) Removal, installation, and repair of
landing gear tires.
(2) Replacing elastic shock absorber cords
on landing gear.
(3) Servicing landing gear shock struts by
adding oil, air, or both.
(4) Servicing landing gear wheel bearings,
such as cleaning and greasing.
(5) Replacing defective safety wiring or
cotter keys.
(6) Lubrication not requiring disassembly
other than removal of nonstructural items
such as cover plates, cowlings, and fairings.
(7) Making simple fabric patches not requiring
rib stitching or the removal of structural
parts or control surfaces. In the case of
balloons, the making of small fabric repairs
to envelopes (as defined in, and in accordance
with, the balloon manufacturers’ instructions)
not requiring load tape repair or
replacement.
(8) Replenishing hydraulic fluid in the hydraulic
reservoir.
(9) Refinishing decorative coating of fuselage,
balloon baskets, wings tail group surfaces
(excluding balanced control surfaces),
fairings, cowlings, landing gear, cabin, or
cockpit interior when removal or disassembly
of any primary structure or operating
system is not required.
(10) Applying preservative or protective
material to components where no disassembly
of any primary structure or operating
system is involved and where such
coating is not prohibited or is not contrary
to good practices.
(11) Repairing upholstery and decorative
furnishings of the cabin, cockpit, or balloon
basket interior when the repairing does not
require disassembly of any primary structure
or operating system or interfere with an
operating system or affect the primary
structure of the aircraft.
(12) Making small simple repairs to fairings,
nonstructural cover plates, cowlings,
and small patches and reinforcements not
changing the contour so as to interfere with
proper air flow.
(13) Replacing side windows where that
work does not interfere with the structure or
any operating system such as controls, electrical
equipment, etc.
(14) Replacing safety belts.
(15) Replacing seats or seat parts with replacement
parts approved for the aircraft,
not involving disassembly of any primary
structure or operating system.
(16) Trouble shooting and repairing broken
circuits in landing light wiring circuits.
(17) Replacing bulbs, reflectors, and lenses
of position and landing lights.
(18) Replacing wheels and skis where no
weight and balance computation is involved.
removal of the propeller or disconnection of
flight controls.
(19) ...
(20) Replacing or cleaning spark plugs and
setting of spark plug gap clearance.
(21) Replacing any hose connection except
hydraulic connections.
(22) Replacing prefabricated fuel lines.
(23) Cleaning or replacing fuel and oil
strainers or filter elements.
(24) Replacing and servicing batteries.
(25) Cleaning of balloon burner pilot and
main nozzles in accordance with the balloon
manufacturer’s instructions.
(26) Replacement or adjustment of nonstructural
standard fasteners incidental to
operations.
(27) The interchange of balloon baskets and
burners on envelopes when the basket or
burner is designated as interchangeable in
the balloon type certificate data and the baskets
and burners are specifically designed for
quick removal and installation.
(28) The installations of anti-misfueling devices
to reduce the diameter of fuel tank
filler openings provided the specific device
has been made a part of the aircraft type
certificiate data by the aircraft manufacturer,
the aircraft manufacturer has provided
FAA-approved instructions for installation
of the specific device, and installation
does not involve the disassembly of the existing
tank filler opening.
(29) Removing, checking, and replacing
magnetic chip detectors.
(30) The inspection and maintenance tasks
prescribed and specifically identified as preventive
maintenance in a primary category
aircraft type certificate or supplemental
type certificate holder’s approved special inspection
and preventive maintenance program
when accomplished on a primary category
aircraft provided:
(i) They are performed by the holder of at
least a private pilot certificate issued under
part 61 who is the registered owner (including
co-owners) of the affected aircraft and
who holds a certificate of competency for the
affected aircraft (1) issued by a school approved
under § 147.21(e) of this chapter; (2)
issued by the holder of the production certificate
for that primary category aircraft
that has a special training program approved
under § 21.24 of this subchapter; or (3) issued
by another entity that has a course approved
by the Administrator; and
(ii) The inspections and maintenance tasks
are performed in accordance with instructions
contained by the special inspection and
preventive maintenance program approved
as part of the aircraft’s type design or supplemental
type design.
(31) Removing and replacing self-contained,
front instrument panel-mounted
navigation and communication devices that
employ tray-mounted connectors that connect
the unit when the unit is installed into
the instrument panel, (excluding automatic
flight control systems, transponders, and
microwave frequency distance measuring
equipment (DME)). The approved unit must
be designed to be readily and repeatedly removed
and replaced, and pertinent instructions
must be provided. Prior to the unit’s
intended use, and operational check must be
performed in accordance with the applicable
sections of part 91 of this chapter.
(32) Updating self-contained, front instrument
panel-mounted Air Traffic Control
(ATC) navigational software data bases (excluding
those of automatic flight control
systems, transponders, and microwave frequency
distance measuring equipment
(DME)) provided no disassembly of the unit
is required and pertinent instructions are
provided. Prior to the unit’s intended use, an
operational check must be performed in accordance
with applicable sections of part 91
of this chapter.

I don't see an exact match, except for possibly "(30) The inspection and maintenance tasks
prescribed and specifically identified as preventive
maintenance in a primary category
aircraft type certificate or supplemental
type certificate holder’s approved special inspection
and preventive maintenance program
when accomplished on a primary category
aircraft provided: ..."  Someone with access to the maintenance manual would have to check to see if it is listed there.

I would have my doubts about that. No maintenance manual for any aircraft that I have ever worked on has ever said that replacing a non-consumable part on an engine is preventive maintenance.

FAR Part 43, Appendix A, tells you exactly what preventive maintenance you can do as the owner of an aircraft that is not used for commercial service.

could you consider quoting here the relevant parts of the FAR?

Just replace the drain at the shop in front of the A&P

Not that simple. The A&P needs to ensure all work is accomplished in accordance with appropriate data and then sign off the work for return to service.

Sure. But this is a question about the owner not the A&P

That's an interesting question. In reality, anyone can replace a part on their own A/C, but to be legal, you must be under the supervision of a certificated A&P and get it inspected/signed off.
FAA, part 43.3 (d)

Welcome to Aviation.SE! Can you provide any sources that back up this answer?

Title 14, part 43.3 (d).

Larry's answer is almost correct. The section he refers to (14 CFR section 43) lists preventive maintenance that the pilot or owner may do and make the appropriate log book entries.  Under this section no A&P supervision is required.  So it is clear that the pilot/owner can change the oil which involves removing the oil drain.  The question is: does this permit him/her to use a different plug as he finishes?  If it is the same part number as the OEM plug I vote yes.

Sorry Skip, The way I read it is:

(d) A person working under the supervision of a holder of a mechanic or repairman certificate may perform the maintenance, preventive maintenance, and alterations that his supervisor is authorized to perform, if the supervisor personally observes the work being done to the extent necessary to ensure that it is being done properly and if the supervisor is readily available, in person, for consultation. However, this paragraph does not authorize the performance of any inspection required by Part 91 or Part 125 of this chapter or any inspection performed after a major repair or alteration.

This, in my view, constitutes an alteration.

Larry, please look at Section 43(g).  A pilot owner may do preventive maintenance without the supervision of an A&P.  It says:(g) Except for holders of a sport pilot certificate, the holder of a pilot certificate issued under part 61 may perform preventive maintenance on any aircraft owned or operated by that pilot which is not used under part 121, 129, or 135 of this chapter. The holder of a sport pilot certificate may perform preventive maintenance on an aircraft owned or operated by that pilot and issued a special airworthiness certificate in the light-sport category.

@Larry whether or not this constitutes an alteration is the heart of the issue.  If you cannot back your opinion on that subject, then this doesn't answer the question.  So far all you've said is "This, in my view, constitutes an alternation."  Can you back it up with any actual references?

Part 43, appendix d. Read the regulation. As soon as the pilot installs the OEM plug he brings the A/C back to Mfg specifications. To install the quick drain is legally an alteration and requires an A&P.

@Larry is a quick drain considered part of the powerplant and subject to approval as a part and or accessory?

Larry...does your last name start with A in sdl near dvt?

Short answer is YES.
Read FAR 43 Appendix A thoroughly and find (23) which says:

(23) Cleaning or replacing fuel and oil strainers or filter elements.

A quick drain valve is considered an oil strainer.
Create a logbook entry and have a beer.

Welcome! Just to play devil's advocate, do you have a FAR reference showing that a quick drain plug is considered an oil strainer?

A drain valve is NOT an oil strainer. An oil strainer is a mechanism contained within the pressure portion of the oil system used to strain oil. It's function is similar to, and usually in place of, that of the oil filter used on many piston aircraft and cars.


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Where is the landing distance for seaplanes specified?
I need to know, is there a specified distance for the landing areas of seaplanes? In airports, we have generally distances are defined according to the flight requirements. I need to know whether there are certain distances defined or just making the landing area by guessing?

Which type of seaplanes exactly?

Cessna Caravan 208

In general, the landing distance is much less than the takeoff distance so it is a non-issue.

Google "Cessna 208 on floats performance" and you'll find lots of info. The first entry is http://www.wipaire.com/aircraft_page/cessna-caravan-series/, which will show you that the takeoff run on water with the small engine is 2341 feet, takeoff over a 50 foot obstacle 3660 feet. As @MikeSowsun said, a water landing run will be much less (they don't even bother to give it) and is a non-issue. My guess would be less than half the takeoff run. At least that was I think true of the old J3 floatplane I used to fly.

@Terry I think you could turn that into an answer. :) Also, please avoid suggesting to "just google it" ;)

@ Terry, is there any major class of plane you haven't flown?

@dalearn Actually, compared to many I have had a very limited experience insofar as classes of aircraft. I don't have multi-engine seaplane, helicopter, glider, airship, or balloon licensing. I have only four type ratings: CE-500, SA-227, B-727, and B-747. I didn't start flying commercially until I was over 30 years old, and getting to the 747 was pure luck.

is there a specified distance for the landing areas of seaplanes?

A cautionary note: As @MikeSowsun said, the required distance to land a seaplane is MUCH shorter than that required to takeoff. For operational planning purposes, what you should be interested in is takeoff distances. My offhand guess is that landing distances are less than half that of takeoff runs. At least that was I think true of the old J3 floatplane I used to fly.
Now I assume you're including floatplanes as seaplanes. There are relatively few hulled seaplanes these days as compared to floatplanes, and compared to the numbers of the wheeled configuration of a given aircraft for which floats are available, the number of floatplanes is relatively small, which may account in part for the POH (pilots' operating handbook) for aircraft usually not providing performance data for float configurations. There are also differences in performance depending on the make and model of the floats being used.
All that said, you can often find performance details for current production floats on a given aircraft on the float manufacturer's website. For example, http://www.wipaire.com/aircraft_page/cessna-caravan-series/ gives that information for the Cessna 208 on Wipaire amphibious floats. In this instance, the takeoff run on water with the Caravan's small engine is 2341 feet, takeoff over a 50 foot obstacle 3660 feet. The same manufacturer's figures for a Cessna 206 on their amphibious floats is 1770 feet takeoff run, 2850 feet over a 50 foot obstacle.

just making the landing area by guessing?

Not by guessing, but looking at the intended area of operation in the light of past experience and judging whether they can takeoff from that area.

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Why do helicopters have such limited range?
Why do rotorcraft like helicopters have such limited range, rarely exceeding 1000 km on a full tank? For example, the range of the Apache is close to 425 km, and MI-26's is 800 km.

Related How does the efficiency of a fixed wing compare to a rotary wing?

I liked your question but the range is different on the Wikipedia articles. For Apache, it is 476 km, and for MI-26, it is 1920 km.

An airplane tricks the air into letting it fly.  A helicopter beats the air into submission.

@Farhan see this

Because there is always someone gorgeous who lives close by who loves helicopter pilots.  Fixed wing pilots need to fly a lot further.

Helicopters are very inefficient in forward flight, for several reasons.

Consider Glauert's high speed assumption, i.e. the helicopter can be represented as an aircraft with a circular wing during high speed flight.  Now, if you know anything about basic fixed-wing performance, you'll realize immediately that this implies an aspect ratio just over 1, which is awfully small.  You may recall that for a fixed-wing aircraft, induced drag is inversely proportional aspect ratio--so a very low aspect ratio means very high induced drag!  Induced drag is normally not a big deal for fixed-wing aircraft at high speed because it's proportional to the square of the lift coefficient (which decreases with speed), but...
In reality, the situation for a helicopter is even worse!  Glauert's assumption assumes we have an ideal rotor, i.e. one that has a uniform induced velocity throughout.  In practice, the inflow distribution over the rotor is highly non-uniform, especially at high speeds where the advancing side of the rotor is in a very different aerodynamic environment than the retreating side.  Even on a very simple level, the "working area" swept on the retreating side is much smaller than that on the advancing side. Since lift needs to be balanced, momentum theory implies a higher induced velocity over the smaller area "worked" by the retreating side, and therefore a much higher induced drag at high speeds than an equivalent fixed wing aircraft.

Rotor wake computation from TsAGI RC-VTOL code (Kritsky, B.S).
As the helicopter goes faster, and as the rotor turns more slowly, more of the retreating side blade will stall, which leads to a dramatic increase in drag. This starts on the inboard section of the blades, and moves outwards as the advance ratio (ratio between flight speed and tip speed) increases.  In fact, the most inboard sections of the blade will even be in reversed flow--and airfoils don't work too well backwards.
On the advancing side, things are bad too, but for different reasons. The profile drag of the rotor increases rapidly with speed.  Moreover, as the advance ratio increases, the advancing tip Mach number also increases--and as the blades get into the high transonic regime, local shocks begin to form and lead to even larger increases in drag.  This can be delayed by slowing the rotor down, but that makes the problems on the retreating side even worse!  In general, it's hard to develop airfoils that work well in both regimes.
Outside of the rotor itself, the rotor hub is very draggy, between the main rotor shaft, pitch links, swashplate, hinges and blade grips, etc.  Helicopters usually have pretty "utilitarian" fuselages, as well, which are not so well streamlined.  Plus, you have the tail rotor, which has most of the same problems as the main rotor.
Back to the specific question of range, the Bregruet range equation says that the range of an aircraft is proportional to the lift-to-drag ratio and the speed of the aircraft.  We've already established that the L/D of a helicopter is pretty bad, especially at speed.  So, the cruise speed is going to be pretty low and the L/D is going to be pretty low, so to sum up, the range is going to be really low.

The tradeoff is that the helicopter is very good at hovering--even amongst VTOL platforms, the helicopter is the most efficient. Speed and range are the price you pay for that capability.

Because they are less efficient than fixed-wing aircraft.

The rotor blades need to move significantly faster than the craft and therefore incur significantly more drag than fixed wings.
More drag is caused by the suboptimal lift distribution. The centre of the rotor creates little lift because the blades are slow there and is shielded by the fuselage, which causes additional vortices in the middle of the downwash.
The rotor diameter is limited by blade tip speed and material strength, which again limits the efficiency, since it can only affect limited amount of air and has to therefore accelerate it to higher speed, which needs more energy.
Aircraft achieve efficiency by flying high, where the air is thinner and thus drag is lower, and compensate the lower lift by flying faster. But for helicopters it is not possible due to physical limits on the rotor size and speed, nor would it really work, because their induced drag does not decrease so much with speed.
Due to all those limitations, helicopters have relatively less lift for their dry weight and therefore can't take as much fuel.

I would add that helicopters are not as efficient from the aerodynamic standpoint. The fuselage is always just under the rotor disk, so will be affected by airflow (unless we do away with the cabin ;-)).

Also: limited fuel tank capacity (2 tons) and more than 10% of the energy lost to power the tail rotor and counter the torque effect created by the main rotor.

@ALANWARD, well, the fuselage is under the centre of the rotor disk where it produces least lift. But than that is part of the problem. For efficiency it would be best if the rotor produced one nice vortex ring, but that would require the centre to produce most lift and it can't do that.

Helicopters don't fly.  They beat the air into submission.  This is generally fuel inefficient.

Basically, for the same gross mass, a rotorcraft requires considerably more power than a fixed-wing. This is due to the fact that fixed wing aircraft propulsion system needs to create thrust that is equal to the drag of the complete aircraft, and the weight of the aircraft is lifted by the wing-lift force (which is usually 10 or 20 times higher than drag); whereas the rotor of a rotorcraft needs to create thrust that is equal to the total weight of the rotorcraft.
Even if both aircraft have the same engine efficiencies, the helicopter fuel-consumption rate (liters per hour) would be considerably more. So, theoritecally helicopters are not comparable to the efficiency of fixed-wing.

Where do you get the 10x power/weight figure from?  In forward flight, a helicopter isn't that much different than a comparable aircraft--the rotor functions more or less like a wing, albeit a poor one.  Looking around for fuel consumption figures, I seem to get ~2x fuel consumption for helicopters compared to comparable prop-driven fixed wing aircraft.   See for instance this Aviation.SE question or this thread on the Straight Dope forums.

@EricGreenwood, you're right that with forward flight the efficiency of the rotorcraft increases, but still it is 2-3 times less efficient than a comparable fixed wing. I've reworded "10 times" to "considerably more". thanks for pointing out.

Would another factor be that helicopters even if helicopters don't need thrust in excess of their weight all the time, they always need to carry around equipment capable of generating it, while fixed-wing aircraft which aren't designed for combat or acrobatics generally don't?

Because in forward flight, the lift/drag ratio of a helicopter is much lower than that of a fixed wing aeroplane.

The rotor forms a circular wing, with an L/D of about 6 at best
A fixed wing aeroplane can have an L/D of over 15.

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For calculating $V_{stall}$, why should be used $C_{L,max}$ and not $C_{L,min}$ (that would be more prudent)?
Premise: In level unaccelerated flight we have relation $W=L=\frac{1}{2}\cdot\rho\cdot V_{stall}^2\cdot S\cdot C_{L,max}$ (taken from book Daniel P Raymer "Aircraft Design: A Conceptual Approach" equation (5.5) at page 85).
Question: Why $C_{L,max}$? Why not $C_{L,min}$? In fact in case the $C_L$ in use is not $C_L,max$, and speed is little more than $V_{stall}$ (valued for $C_{L,max}$), it happen that Lift is not enough and anyway stall happens on aircraft. If using $C_{L,min}$ the calculation is more cautious.

What do you mean by $C_{L,min}$? $C_{L,max}$ is the maximum coefficient of lift the wing can provide, that is coefficient of lift at stall $\alpha$. At other angle of attack it can provide less lift down to and including none at all. So what would $C_{L,min}$ mean?

This is not off topic, calculating stall speed is extremely important to flying certain types.  Why is there a close vote on this?

The stall speed is the speed at which you can still fly the plane in level flight. This is important, because strictly speaking, stall happens at a certain angle of attack, not at a certain speed. In other words, you can safely fly a plane way below stall speed - the problem is that you'll be doing it nose-down in order to keep the angle of attack to a minimum (and that as such you will quickly exceed the stall speed again - hey, we just did a stall recovery maneuver!)
So, let's see what happens when we approach stall speed from a higher speed, while flying level. As we slow down, we must increase our angle of attack to maintain level flight - this way, we are increasing the $C_L$ of the wing, as to keep the lift $L(=W)=\frac{1}{2}\rho VSC_L$ constant when reducing our $V$. Ideally, we'd like to keep increasing our $C_L$ as we reduce our $V$ - however, this means we keep increasing our AoA (angle of attack) until it gets past a critical point, where the airfoil stalls due to flow separation at the suction side of the airfoil.
So really, using $C_{L,min}$ would be nonsensical - if your wings would have a lift coefficient so low at such a low speed, I can guarantee you that $W>L$, i.e., you plummet from the skies. We need to increase our $C_L$ to maintain level flight, and the stall speed is exactly the point where we can no longer increase our $C_L$, or in other words, we arrived exactly at $C_{L,max}$.

The general equation during level unaccelerated flight is $L=\frac{1}{2}\cdot\rho\cdot V^2\cdot S\cdot C_{L}$:

For a variety of speeds V between the stall speed and the Never Exceed speed.
For a variety of altitudes, between under sea level and the stratosphere.
For a variety of Angle of Attack of the aeroplane, equating to a range of $C_L$
Resulting in a variety of lift forces L.

There are three independent variables in the one equation, chose all three and the lift will be found, choose the lift and two others and the third one will be found.
This latter thing is what Raymer is doing. He sets lift to weight W for steady unaccelerated flight; he sets $C_L$ to $C_{L,max}$; and he then finds the minimum speed $V_{min}$ that the aeroplane can fly with at a certain altitude.
Note that for normal level unaccelerated flight $C_L$ is always > 0. The minimum $C_L$ is a negative value for negative AoA, and would mean that the aeroplane is flying upside down.

There are multiple stall speeds that have been defined. For example, the 14 CFR §1.2 Abbreviations and symbols lists the following stall speeds:

$V_{S}$ means the stalling speed or the minimum steady flight speed at which the airplane is controllable.
$V_{SO}$ means the stalling speed or the minimum steady flight speed in the landing configuration.
$V_{S1}$ means the stalling speed or the minimum steady flight speed obtained in a specific configuration.
$V_{SR}$ means reference stall speed.
$V_{SRO}$ means reference stall speed in the landing configuration.
$V_{SR1}$ means reference stall speed in a specific configuration.
$V_{SW}$ means speed at which onset of natural or artificial stall warning occurs.

What you're asking is the $V_{S}$, the minimum flight speed at which the airplane is controllable. In short, as $V_{stall} \propto \sqrt{\frac{1}{C_{L}}}$, $C_{L_{max}}$ gives the minimum speed at which the aicraft is controllable.
Consider an aircraft in a level flight, If the pilot wants to reduce the speed, in order to maintain a steady, level flight, he/she has to increase the angle of attack i.e. increase the $C_{L}$. He can do this till the $C_{L}$ reaches $C_{L_{max}}$, where the speed becomes minimum while the aircraft is still in a steady, level flight. If the speed is reduced any further, the aircraft loses lift; this speed gives the stall speed of the aircraft.
You can get the stall speed for any configuration by using the $C_{L}$ at that configuration; but the values have to be realistic. Setting the $C_{L}$ to very low values to get large $V_{stall}$ makes no physical or practical sense. For example, the aircraft can be set to have zero (or even negative) $C_{L_{min}}$, in which case, the stall speed has no meaning.

In the equation $W=L=\frac{1}{2}\cdot\rho\cdot V_{stall}^2\cdot S\cdot C_{L,max}$ the goal is to find minimum flight speed which can produce lift equal to weight. This speed varies in different configuration but the minimum would be at maximum angle of attack producing $C_{L,max}$ (lift coefficient). The lift coefficient is a function of angle of attack, airfoil and wing geometry.

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What is the significance of stall speed
Irrespective of the stall speed, the only reason aircraft stalls is when its control surface (elevator, aileron) exceeds critical AOA. Then what is the significance of the stall speed?
Does the aircraft stall at the "stall speed" even if its AOA is less than critical?

"the only reason aircraft stalls is when its control surface (elevator, aileron) exceeds critical AOA" This is completely wrong, stalls have nothing to do with the elevators or ailerons.

See also: http://aviation.stackexchange.com/questions/14909/why-is-stall-speed-listed-in-a-poh

The word "stall" in general means the aircraft can no longer fly. Now, the reason for an aircraft to stall can either be that the wing can no longer generate adequate lift due to excessive flow separation (which, confusingly is also called a "stall" condition) or that the wing can no longer generate adequate lift because there is not enough air blowing over it (the aircraft is too slow). This is what is meant by stall speed when quoted by manufacturers: don't fly your plane slower than this speed. But be aware that you can also stall at almost any speed.

The "stall speed" should be renamed "the speed below which the aircraft cannot be maintained at the same altitude without exceeding the stall angle". It's a shortcut. When you slow the aircraft, the AoA must be increased to maintain the altitude, but at some point this is not possible any more, the aircraft stalls if the flight is maintained horizontal.

There is not one stall speed, instead there are several, depending on aircraft configuration and flight conditions. They are used to determine speeds for the safe operation of the aircraft or for limit speeds.
For example: The stall speed in landing configuration and level flight with the actual weight is used for determining the approach speed. By multiplying with a safety factor of 1.3, you arrive at the recommended approach speed. Or: The stall speed in take-off configuration is used for determining the rotation speed. And so on.
Another example: The speed range in which it must be safe to deploy flaps is limited by 1.4 times stall speed of the clean aircraft, or 1.8 times the stall speed in landing configuration, whichever is greater. Or: The maneuvering speed v$_A$ is calculated by multiplying the stall speed of the clean aircraft with the square root of the maximum allowable load factor. And so on.
If you wonder what a stall is, maybe it helps to read the answers to this question. It is not flow separation or exceeding a specific angle on the control surfaces that determines the stall condition, but the reversal of the lift curve slope over angle of attack: Stall is when more angle of attack will no longer increase lift, so it is the angle of attack of the whole aircraft that counts. Physically, this means that large-scale flow separation appears over part of the wing. A good design will still have fully working control surfaces at this point.

No. If the angle of attack is not critical, then it does not matter what the relative air speed is, the plane will not stall. By definition the "critical" AOA is the angle the plane stalls at, so you are asking sort of a tautological question.
The stall speed is there for guidance only, to give the pilot a general idea of at about what point the aircraft will begin to stall in level flight.
In practice the speed at which the aircraft actually stalls will vary according to a lot of different conditions such as temperature, humidity, angle of bank, gust conditions and many other factors, even how clean the wings are.

"By definition the "critical" AOA is the angle the plane stalls at" Not the plane, the airfoil. The angle of the aircraft is different depending on the wings and the AOA changes along the length of some wings that have a twist in them for any given pitch angle.

@RonBeyer True, but I think you are kind of splitting hairs. The OP does not seem to be at the point where knowing the details of airfoil and wing geometry would be relevant.

Either way I think the goal should be the accuracy of the answer, an airplane can stall at pretty much any pitch angle, but there is a critical AOA where the airfoil will stall, and the two are not interchangeable. Lets not add confusion by associating pitch angle with AOA.

@RonBeyer Who said anything about pitch (besides you)? Neither me nor the OP said anything about pitch.

Angle of the plane = pitch...

@RonBeyer Neither myself nor the OP referred to either the "angle of the plane" or to "pitch".

"By definition the "critical" AOA is the angle the plane stalls at" ??

@RonBeyer Yes, the critical angle of attack is angle at which the plane stalls. By that I was referring to the AOA, not the pitch.

authors: Виктор Захаров


text:
Why does this analysis show very high stress in a propeller?
I have been trying to evaluate the stress on aluminium propeller:
Diameter: 78", 2600RPM
Have fed load numbers from here, applied distributed mass model onto the blades.
Static Thrust (2600 rpm):
MT 2-blade, 83" composite, 920 pounds
MT 3-blade, 80" composite, 900 pounds
McCauley 2-blade, 82" aluminium, 860 pounds
McCauley 3-blade, 78" aluminium, 825 pounds

The results are rather scary - 30 MPa out of 55 max, for 6061 class Aluminium (the visual deformation is increased 20x for better representation),
resulting in +-1cm vertical bending.
Is this normal?
Is there any info on industry's practice for safety margin (numbers show its less that 50%)? Or any kind of test procedures applicable to manufacturers?

The shear stresses are negligible - 150 Pa out of 150 MPa

The two propeller case is really scary looking
Below is a distributed mass (375Kg) - 80% of Max. yield stress

And centrifugal force - over-stressed by 200%:

Finally, a combined non-Linear analysis (incl. torque) - over-stressed by 250%:
(notice that everything not in blue will bend and likely crack, with only the blade tips surviving the torture)

Apart from the thrust load, did you take into account the centrifugal force...?

I didn't consider it, and yes, intution seems to propose that a propeller should strenghten itself in XY plane. I edited the post to include it. It has made things even worse.

Are you sure you are using the correct blade profile in your calculations?  GIGO

The static case may not even be limiting. Vibration and fatigue are probably more critical and are covered by 14 CFR §25.907. There are also acceleration loads.

I know this sounds silly, but perhaps some of the numbers in the calculation are off by a magnitude or the unit is wrong?

If you are looking for some useful feedback, it would be handy for you to post your full set of applied boundary conditions and mesh (inc screenshots). There are indications here to me that you have incorrectly restrained your model; and that you have possibly copied your geometry from a wind turbine instead of an aeroplane propellor. The two function in different ways and cannot be transposed from one function to the other.
Also, your scales need some work. Standardise the scales to a fixed range instead of whatever is the max experienced.

I fixed your units (material science has along way to go before a yield strength of 5.5GPa can be achieved in aluminium). It would be advisable to get familiar with common values for stresses and moduli in SI units. @kevin I suppose that made your comment obsolete; 30MPa stress seems perfectly realistic for the loadcases mentioned.

Good catch. My sincere apologies for this brutal mistake. However, that has made me wonder if 6-series Aluminium was the correct choice. 7-series would withstand 103MPa at a cost of complicated extrusion, which wouldn't be used anyway.

The plots of your propellers seem to use a regular airfoil all the way to the root. Real propellers become almost cylindrical there, however. Also, aluminium propellers are hollow, so the outer part is only a thin shell, saving mass and reducing centrifugal loads significantly.
After all, real aluminium propellers work, so there must be a way to keep the stress in operation below the yield stress. FAR 23.907(c) even demands that a non-wooden propeller must be life-safe:

The applicant must perform an evaluation of the propeller to show that
  failure due to fatigue will be avoided throughout the operational life
  of the propeller using the fatigue and structural data obtained in
  accordance with part 35 of this chapter and the vibration data
  obtained from compliance with paragraph (a) of this section.

Below is a drawing of a wooden propeller, but the outer shape of a metal propeller is not much different.

Drawing of a wooden propeller for the Fokker D VII (picture source)

Section 35 (AIRWORTHINESS STANDARDS: PROPELLERS) is exactly what I have been looking for.
I will try to run the similar tests on some WW2 propellers.

@FlegmatoidZoid: Cool, then please let us know what results you get.

For the design of propeller consider 1.5 times of rpm. This assumption gives double the centrifugal force, thrust and drag acting in the propeller.
Consider the endurance limit based on following data. This helps to avoid fatigue problems

If you could indicate the source of your data, that would increase the credibility of your answer significantly!

the exact value is 1.41.   Apply the value in equation to understand the changes.


text:
Am I required to report entering the hold in lieu, persuant to the IFR mandatory reporting points?
You are cleared for the RNAV (GPS)-A approach at 0S9.
You are cleared direct JIGEB then at or above 3200 to YAYUB then cleared RNAV GPS A.  You are not asked to report anything.  Upon arriving at YAYUB you perform a parallel entry to the hold reverse course and the proceed inbound.
If this was a procedure turn I am not required to report the procedure turn unless ATC requests it.  However, this is a hold in lieu of a procedure turn.  Am I required to report entering the hold persuant to the IFR mandatory reporting points?

If you're in radar contact with ATC, the answer is no, you will not need to report it unless requested to.  If you are not in radar contact (which would be unusual as that is in Class E airspace) you would not have to report your position entering the hold at YAYUB unless you were instructed to hold at this point with an EFC time.  From the AIM 5-4-9

When a holding pattern replaces a procedure turn, the holding pattern must be followed, except when radar vectoring is provided or when NoPT is shown on the approach course. The recommended entry procedures will ensure the aircraft remains within the holding pattern’s protected airspace. As in the procedure turn, the descent from the minimum holding pattern altitude to the final approach fix altitude (when lower) may not commence until the aircraft is established on the inbound course. Where a holding pattern is established in-lieu-of a procedure turn, the maximum holding pattern airspeeds apply.

But 5-4-9 also states that

If cleared for the approach prior to returning to the holding fix, and the aircraft is at the prescribed altitude, additional circuits of the holding pattern are not necessary nor expected by ATC. If pilots elect to make additional circuits to lose excessive altitude or to become better established on course, it is their responsibility to so advise ATC upon receipt of their approach clearance.

You would have to report leaving YAYUB inbound and to report crossing WIKUV FAF inbound as well.


text:
Are there cases where a double horizontal stabilizer is useful?
I was watching Agents of Shield recently, and in it the team fly a plane that's claimed to be a modified C-17 Globemaster. It has two extra engines, rotary turrets, and, according to the wiki, a pair of half-length wings at the rear of the fuselage, seen here:
(image courtesy of Agents of SHIELD wiki)
To me, it also kind of looks like a second horizontal stabilizer. Certainly some of the modifications to this fictional plane are pure flights of fancy, but is there any basis behind this one? Is there any reason why this feature would be useful? Have there possibly been any experimental or production aircraft with something like this? Or is this pure "cool factor," with no real basis in aviation?

putting an engine right in the exhaust of another engine seems like a bad idea to me...

there's also shock bodies on the rear wing

it apparently has VTOL capability as well, by rotating the engines forward, I guess that's why there needed to be an engine in the back.

@ratchetfreak couldn't they just mount the rear engines straight on the fuselage?

@raptortech97 then they wouldn't have room for the cargo hatch

See this question for more biplane tails.

The main aerodynamic purpose of the horizontal stab (or certain canards) is to provide longitudinal stability.
If the rear wing with the 5th and 6th engine flies "up," like the main wing, then it will counteract the longitudinal stability of the horizontal stabilizer. If the rear wing flies down, like the h-stab, then it is just extraneous, since the h-stab can be made as large as necessary.
If all you want to do is to add extra engines, they can be mounted on the main wing, like in the eight-engine B-52 bomber:

or the six-engine Antonov An-225 cargo jet:

a joke> A reader wrote us, retelling the story about the military pilot calling ATC for a priority landing because his single-engine jet fighter was running "a bit peaked." ATC told the fighter jock that he was number two behind a B-52 that had one shut down. "Ah," the pilot remarked, "the dreaded seven-engine approach!"

Oh, but this question was posted right after the aircraft was initially seen on the show. In a later episode, we discover the aft positioning of the extra engines is essential.  Spoiler alert!

@rbp has a good answer.  I'd like to add something to it.
For most aircraft, the horizontal stab provides stability and drag but relatively little lift.  The wing provides 100% of the lift and everything else out there provides stability.  And, of course, everything in the airflow (aside from the engines, which provide thrust) provides drag.
Contrast that with a canard, which provides stability and lift (and some drag, coincident with the lift).  The canard typically provides 10 - 20% of the lift, with the main wing providing the rest.  By putting the canard in front, and designing it to stall before the main wing does, the canard will be unable to lift the nose high enough to cause the main wing to stall.  It's not 100% safe; there are still cases where a canard aircraft can stall but they're really obscure.  Dick Rutan, who served as test pilot for Burt Rutan's canard-based aircraft designs, once joked that he could take one of Burt's planes up and try like to crazy to make it stall but "no joy; all I ever got was exercise."
Back in the late 1980s, Airbus started designing the tailplane to provide significant lift.  After takeoff, they shift some of the weight aft (usually by moving fuel around) and take advantage of that.  Airbus has been using this for over a decade to achieve greater fuel efficiency from their aircraft.  With improved fly-by-wire flight controls, they've gotten to the point where they don't have to wait until after takeoff.  The C-17 uses this idea, too (including the fly-by-wire).  But it's my understanding that the tailplane provides no more than 10% of the lift of the main wing.
For the fictional aircraft, they wanted the aircraft to be able to hover.  So they have main engines which can pivot downward.  When in a hover, though, you need some lift forward and aft of the center of gravity (CG) to provide forward / aft stability and translation.  Putting engines on the tail provides that.  Making the tailplane a 1/2 span wing, with appropriate amount of lift, guarantees that approx 1/3 of the total lift will be provided by the tailplane.  Which means that, when the engines pivot downward to hover, 2/3 of the total hovering lift is provided by the wing and 1/3 is provided by the tailplane.  In this fashion, the plane is balanced in normal flight and in hovering flight.
I would've liked to see a large aircraft with the main wing aft of CG and canards, with engines on the canards (or maybe fuselage mounted, near the canards).  But they went with a more "familiar" look; there's no heavy-lift aircraft out there with canards like I'm describing.  The Tu-144 and Valkyrie both have canards, but nowhere near that large.
An F-35 has the tail of the engine pivot downward, aft of CG, and has a "lift fan" forward of the CG.  A Harrier has a total of 4 downward columns of air coming off the engine when hovering, two forward of CG, two aft.

the horizontal stab provides stability and drag but relatively little lift Actually it provides negative lift, which is how it provides stability. Do you have a source for getting 10% lift from a horizontal stabilizer?

@fooot: Being subject to AoA changes, the amount of downforce at the tail varies and might even change to lift at high AoA (depending on cg location). With relaxed stability, a lift-producing tail is normal (on gliders, for example) and still provides stability. With about 15% of the area of the wing, 10% as the upper limit of lift on a normal tail and with relaxed static stability sounds reasonable.

@PeterKämpf I won't disagree that it can vary, or provide lift for applications like gliders, but does this also apply to airliners in cruise?

@fooot: No, the tails of airliners in cruise normally produce a downforce, unless the cg is shifted back like in some Airbus types. This can be seen by the negative camber of the horizontal tail.

My understanding: the horiz stab is normally designed to provide drag and no lift; this provides static longitudinal stability  Drop the elevators and the tailplane provides just enough lift to raise the tail.  Actively flying the elevator in a neutral condition, with less h stab drag (by design), would require a very light touch and very active control; very tiring for the pilot.  That would require more dynamic stability.  A FBW system can provide that without exhausting the pilot.  Most aircraft aren't FBW, so most aircraft are designed for greater static stability.  Modern Airbus are FBW.

Do they have a nefarious super-weapon in the rear of the fuselage? That could necessitate the extra lift and, more importantly, the shifting of that lift further back to keep the centre of lift near the centre of gravity. That would also explain the extra engines, as the extra power is needed.
It also needs the extra engines to maintain the pretence of balance when performing VTOL maneuvers (although this would again suggest the weight has been heavily shifted backwards behind the "regular" wing)
Overall, though, it's a bad design which wouldn't be the ideal way to handle any of those things - it's mostly for cool-factor.

There's no nefarious super-weapon, just a couple of cars.

Double horizontal stabiliser have been recently introduced also in the helicopter world with the H160 but for a slightly different reason.

Horizontal stabiliser on the H160. Source https://en.m.wikipedia.org/wiki/Airbus_Helicopters_H160#/media/File%3AH160_Tail.JPG
Horizontal stabiliser in helicopters mainly provides pitch stabilisation like in a conventional airplane plus some other useful function like, for example, limiting the fuselage's pitch attitude in forward flight, giving less drag.
The standard position of the horizontal stabiliser is at the far end of the tailboom since this position maximize its structural and aerodynamical characteristics; anyway it has also some drawback. In hover, the wake of the main rotor doesn't reach the stabiliser since it contracts and moves downward:

Main rotor wake in hover. The horizontal stabiliser would be at the far right of the tailboom, just before the tail rotor and out of the wake.
However, as the forward speed increases, the wake is more and more skewed backward and at a certain speed it impinges on the horizontal stabiliser:

Main rotor wake in forward flight. Both images from this presentation which, in turn, uses images from J. G. Leishman, Principles of helicopter aerodynamics.
At that point the stabiliser, which is already generating a stabilising downward lift, generates an even bigger download which makes the fuselage more or less suddenly pitch up. The opposite is true when the helicopter slows down and the stabiliser exits the wake. This phenomenon is called... well, pitch-up.
So, a horizontal stabiliser should have:

a big surface to generate enough lift to stabilise the helicopter but
a small (ideally zero) surface as well to minimise the pitch-up.

A smart solution to this contradiction come from noticing that:

the stabilising lift is proportional to the (mainly horizontal) flying speed while
the pitch-up force is proportional to the (mainly vertical) main rotor wake.

Splitting the stabiliser surface into a biplane configuration solves this contradiction: the stabilising force in pitch is unchanged since the total stabilising lift is the same but the rotor wake sees now just the upper half of the stabiliser since the lower half is in the shadow of the upper one. Smart indeed.

To add to the answers of @Meower68 and @rbp The craft's configuration is basically a conventional wing with a biplane tail. Such tails date from the pioneer days of aviation. The exact variant here is a cantilevered inverted seqsuiplane; both cantilever biplanes and inverted sesquiplanes have since flown successfully.
Another concept dating from those days is the lifting tail. Provided it does not work so hard as to destabilize the plane, a lifting tail can usefully offload the wing. A lifting lower stabilizer would certainly have been useful in the fictional design to carry the weight of the extra engines without upsetting the trim.
Someone said that the engines move forward when they pivot to vertical mode. Extra engines in an aft location would indeed be necessary to maintain trim then, as well.
All in all the design is mad, but not barking mad.

A biplane elevator was useful in improving the arc of fire for a gunner, where the gunner is located near the wing - such as with the Hannover CL.II, III and IV, but it was a very rare solution, and even on their CL.V they abandoned it.

On a transport aircraft, it might result in fewer collisions between ground vehicles and the tail, but offers no flight or structural advantages.
In any case, it would result in more drag and weight and less aerodynamic efficiency. Any time an airfoil is superimposed over another, the low pressure area above, and the high pressure area underneath interfere, reducing lift dramatically, while also increasing drag. Staggering them doesn't help much, and the gap necessary to reduce this needs to be huge compared to the chord of the airfoils. Worse, it also adds two extra tips, where vortices are generated when the high and low pressure air spills over the ends, and mixes, hence why in many cases the rudders are placed to act and end plates.

My only thought is that if that rear wing is truly generating a lot of lift, and if the front is more bulbous (and thus more heavy) that plane is just going to pitch forward in flight (if it ever got off the ground) and go tail over head until it crashes.
Basically, the center of lift (between the two wings) is going to be way behind the center of mass (well in front of the wings), and that usually does not end well.

authors: محمود العراقي


text:
People say that modern airliners are more resilient to turbulence, but I see that a 707 and a 787 still have the same G-rating. Why is this?
The 707 was introduced in the late 50's, the 787 in 2009.
Back when Air France 447 had recently gone missing, many were speculating that the aircraft had broken up in the turbulence associated with the massive thunderstorm it was flying through. Some aviation spokesperson, who I believe were associated with Airbus, said that the planes are designed to withstand the worst storms with a 50% safety margin.
Several airliner models from the 50's and 60's have broke apart in flight due to turbulence associated with thunderstorms (A 720 B, A BAC-1-11, and a F-27 in 1981). There were also breakups involving clear-air turbulence (707, Mt. Fuji, 1966, and a Consolidated Airlines F-27). I checked to see if their G rating was lower than the planes of today and they were the same - 2.5+ G Limit Load, with a 50% safety margin, so 3.75+ G Ultimate Load (beyond which structural failure occurs).
Why is this? Shouldn't the G rating be much higher if modern airliners are more resistant to violent turbulence?

It may have to do with the manufacturer publishing upto the certification requirements, and not capabilities beyond that. Just a thought.

Related question: How come the DC-8 plane can withstand +15G from turbulence, but newer planes can only withstand 20% of that amount?

Why has maximum safe G load not increased in 70 years of jetliner passenger service?  My guess is because with a force of more than 4 G there's a very high potential for the passengers and crew to be killed or injured.  Even if the plane can still fly the problem will be the people inside will have so many broken bones and other injuries that the probability that the people will survive long enough for a landing at an airport is quite small.  Astronauts will see more G forces but they are strapped tight in special seats, in peak physical condition, and trained on how to react.  Just my guess.

Note that the 2.5 g limit is for load due to manoeuvring, not only for turbulence. The certification specification for large aircraft on the subject of turbulence and gusts has changed several times since the certification of the Boeing 707.
Turbulence, and its effects on aircraft, is nowadays much better understood. Different aircraft react differently to the same turbulence. What will result in a high g-load in one aircraft, may be a moderate g-load in another. Therefore the certification regulations now focus on prescribing the type of turbulence that needs to be safely dealt with, instead of the resulting load factor.
In 1964, a formula was introduced describing the gust load that the structure aircraft has to deal with. This was added as FAR 25.341. This section has subsequently been updated in 1990, 1996 and 2015.

Modified from my earlier answer to a related question

Correct me if I'm wrong, but it seems as though the gust load limits are very similar, from 56 fps to 66 fps. This does not seem like a big improvement. That's a  difference of 7 miles per hour. Am I reading the papers wrong?

@LarsKnowles, the original criteria just had a single requirement: handling of discrete vertical gusts.  The 2015 update has three: discrete gusts (both vertical and lateral), continuous turbulence, and paired vertical/lateral discrete gusts.

In addition to Mark's comments, which excellently addresses the main point of my answer, the extra 10 fps in discrete vertical gusts all but insignificant. It's an increase of 18% in speed, and 39% in kinetic energy.

Thanks for the clarification. However thunderstorms and wind around mountains can produce gusts stronger than 66 fps often, can they not?

Resilient to turbulence could also mean that they don’t experience as much roughness when flying through turbulent air. The wing loading of the aircraft has much to do with this. Modern aircraft with more efficient wings and high loading ratios (kg/m2) will be ‘bashed about’ less than older aircraft with lower ratios.
From your example the Fokker F27 has a ratio of 282kg/m2 of wing whereas a Boeing 739 or Airbus A380 have ratios somewhere up in the 689-690kg/m2 region.
Ref

It depends on the interpretation of ‘resistant’.  You’d home that problems like metal fatigue are better understood these days, and modern designs may deal with turbulence without exerting large g-forces.  It’s difficult to imagine passengers being too happy with exposure to 2.5G so avoiding or mitigating such events has to be better than just making aircraft stronger.

Now that we understand metal fatigue, we're building everything out of composites...

Metal fatigue is not a factor in once in a blue moon gust events.  It requires long term cyclic loading, within the load limits.  A gust event that exceeds yield load limits will leave the airplane bent.  If it doesn't exceed yield limits it won't be bent and is good to go. It only comes apart if it exceeds ultimate limits, or if the structure was weakened by fatigue related to normal long term cyclic loading.

You don’t think that pre-existing fatigue might make an airframe more likely to fail when subjected to a gust event?

@Frog Metal fatigue, if it exists under normal loading conditions, will eventually cause the component to fail completely under normal loading conditions.  That is, for example, what causes engines to fail catastrophically once in a while - an undetected manufacturing flaw creates a tiny crack through metal fatigue, which slowly propagates until the fan blade or disc is unable to withstand the forces it is routinely subject to.  But there is so much safety margin in main structural parts that metal fatigue is not a factor there.

It follows that at some point between ‘new’ and ‘broken’ there will exist a state where a 2.5G loading would cause failure, does it not?


text:
Why we don't have to apply cold temperature correction on final approach segment heights (except DA/MDA)?
As shown below, ICAO doc 8168 (PANS OPS) says that cold temperature correction have to be applied on initial and intermediate segments and on the DA and MDA (and missed approache altitudes). But why final approach segment is not mentioned here? We don't have to apply correctons for FAF/FAP altitude?

1.7 COLD TEMPERATURE CORRECTION
1.7.1 Temperatures lower than those of the standard atmosphere result in the actual altitude of an aircraft being  lower than that indicated
by the barometric altimeter. As a consequence, the MOC actually
achieved could be lower  than the prescribed MOC. In order to prevent
this, the pilot shall correct for low temperatures. The pilot is
responsible  for any necessary cold temperature corrections to all
published minimum altitudes/heights in both conventional and  PBN
procedures. This includes:
a) the altitudes/heights for the initial
and intermediate segment(s);
b) the decision altitude/height (DA/H)
or minimum descent altitude/height (MDA/H); and
c) subsequent missed
approach altitudes/heights.

The final approach segment is designed to be safeguarded against the effects of low temperatures (some times a minimum temperature applies)
The explanation is actually given just below the paragraph you quoted from doc 8168, in the section 1.7.2:

1.7.2 The final approach path vertical path angle (VPA) on a 3D approach operation which is based on  barometric vertical navigation
(baro-VNAV) criteria is safeguarded against the effects of low
temperature by the design  of the procedure. This will ensure that the
effective VPA at the minimum temperature published on the chart will
not be  less than 2.5° and has been obstacle assessed. By applying the
cold temperature correction to this procedure type the  nominal VPA
will actually be flown. This can be achieved by manual application of
the correction by the pilot, or in  some cases where certified systems
are used, through automatic application of the correction by an FMS.
Note.— For more information on the use of automated systems for temperature compensation, see the  Performance-based Navigation (PBN)
Manual (Doc 9613).

@757toga Good point. The PANS OPS doesn't say anything about non-precision approaches. You have any idea why?

@757toga I considered that as well, but what about step down approaches that have altitude constraints after the FAF such as in this question. Although that question relates to a US approach, perhaps this there is a difference in this regard between PANS OPS and TERPS


text:
What is the relation between altitude and vertical speed?
I recently saw 2 graphs on a flight path on BBC, shown below. For those interested, it's the path of the recently crashed Boeing 737 max flight ET302, but that's irrelevant to my question.
My question is about the relationship between the two graphs. I would expect the bottom graph to be the first derivative of the top one - with altitude increasing when vertical speed is positive and decreasing when it is negative. However, this does not entirely hold. At around 5:38:45 for example, the altitude clearly drops, while the vertical speed remains non-negative.
Am I missing something about the definitions? Like, vertical speed being defined relative to the wind and the plane stuck in a down-gust. Or, if or example, is there some well-known error caused by the way these quantities are measured by the instruments?

Maybe the altitude on the first graph is relative to the ground (AGL), in witch case the topography of the terrain alone would show as a change of altitude.

@Jimy, no. the data comes from https://twitter.com/flightradar24/status/1104676048317362177 and it is clear that all numbers are w.r.t. sea level

@Federico Oh, my bad, didn't saw where the data came from...


text:
What is the highest a helicopter can fly?
What is the highest theoretical and current flight a helicopter can achieve?

This (and more) is quite easily answered by Wikipedia
Highest altitude without payload
12,442 m (40,820 ft)
Aerospatiale Lama
Highest level flight altitude
11,010 m (36,120 ft)
Sikorsky CH-54 Tarhe
Altitude with 40-tonne payload
2,255 m (7,398 ft)
Mil V-12
Visit the linked Wikipedia article for more records and sources.


text:
Lights at the end of both wings of any airplane
Can anyone please explain to me why there is one green light at the end of wing and on the other side there is red light? what is the purpose of it and why it uses different colors like red and green instead of same color light?
The question is not about the position of the lights but why they use different color lights? like red light on the left side and green light on the right side. Is there any specific reason for that?

https://en.wikipedia.org/wiki/Navigation_light

Same as on ships, the navigation lights (green on right side, red on left side, white on aft tip) are there to allow visual judgement about another vehicle‘s position and orientation (and thus relative headings) at night. E.g:

if you see red and green, you are looking head on at another vehicle
if you see only green, you are looking at the right hand sector of the other vehicle
if you see red and white, you are looking at the left hand aft quarter

and so on.

It makes sense. Thanks for answer.

The green and red lights are called as navigation lights (NAV Lights). It helps for visual judgement of a direction of an aircraft.
If you can see green on your left and red on your right, then it means that the aircraft is heading towards you and vice versa.


text:
Keeping the engine under the wings, doesn't hurt the lift as the airflow speed is higher under the wings?
To create a lift, the airflow speed on top of the wings should be higher than the airflow speed on the bottom of the wings.
But when you keep the engine on the bottom of the wings, wouldn't it hurt the lift as air flows faster on the bottom of the wings?

To create a lift, the airflow speed on top of the wings should be higher than the airflow speed on the bottom of the wings.

No, that's not true.
In order to create lift, the pressure on top of the wings must be lower than the pressure on the bottom of the wings. The airflow speed doesn't matter.
I'm guessing your thought process is something like this:

To create lift, the pressure on top of the wings must be lower and the pressure on the bottom must be higher.
According to Bernoulli's principle, faster moving air has lower pressure and slower moving air has higher pressure.
Therefore, to create lift, the air speed on top must be faster and the air speed on the bottom must be slower.

However, point 2 here is wrong. Bernoulli's principle does not say that faster moving air has lower pressure. What Bernoulli's principle says is that given two points a and b, if

the fluid flow is steady, and
the fluid flow is incompressible, and
viscosity is negligible, and
gravity is negligible, and
the points a and b are on the same streamline, and
the fluid is moving faster at a than at b,

then the fluid has a lower pressure at a than at b.
Since the air above the wing and the air below the wing are not on the same streamline, Bernoulli's principle doesn't tell you anything about the speed of the air going above and below.

In figures posted here, https://i.imgur.com/01yDUem.png, which one is a better design. In Fig.2, the air blown by the engine is passing through a tunnel. Whereas in Fig. 1, the air blown by the engine is directly touching the bottom of wings as it moves back

Well, just because two fluid particles are not on the same streamline doesn't mean Bernoulli can't be applied here. If there's no energy/momentum source (and assume all the other stuff you've mentioned), then the total pressure at far field must be equal for the two particles at two streamlines. If it gains static pressure at some location along the streamline, then its speed must decrease, and vice versa. The proportion of the change must hold for both streamlines.

You're going by an old outdated lift theory, still taught in a lot of places.  The wing induces a very large package of air to move down as it's going along, and this newtonian action/reaction of this package of air being induced to move down is most of "lift".  The Bernoulli part is important, because the pressure differential is a factor and it also is part of what encourages the air above the wing to move.
Anyway, most of the air mass that is motivated to move downward is above the wing.  You can put stuff under the wing and it has little effect on this, but put stuff above the wing and the whole process of making a large package of air extending half a span above the wing to move down is disrupted.
This is why you see airplanes like Skyraiders completely festooned with junk underneath the wing, but which has little effect on the wing's lifting ability.  Put all that stuff above the wing, and it'd never get off the ground.  Same deal with engines.

What I am trying to say is that the air is blown by the engine is touching the bottom of the wings and moving with high speed. Wouldn't it impact the lift?

The propeller wash is adding energy to the entire lift generation process above and below the wing so lift is increased in the area behind the prop.  On a turboprop with 12 or 14 foot propellers, there is a huge effect and power changes have a near instantaneous effect on sink rate.  On a single the effect is there but less pronounced.  It's one of the reasons power on stall speed is less than power off.

Thank you John for your response. Unfortunately, I still did not understand how keeping the engine under the wings helps the lift. Can you post some diagrams describing airflow?

My point was that the velocity of the air above and below and the resulting Bernoulli effect isn't how most lift is created.  It's the large flow field below and mostly above being redirected that creates the lifting force. Having a jet engine under the wing doesn't help or hurt with lift significantly.  Putting the engine under is a better location because putting it, or any other body, on top is very disruptive to the upper flow field.

In figures posted here, https://i.imgur.com/01yDUem.png, which one is a better design. In Fig.2, the air blown by the engine is passing through a tunnel. Whereas in Fig. 1, the air blown by the engine is directly touching the bottom of wings as it moves back.

You want a short tail pipe. Early jet airliners put engines buried in the wings with a long tail pipe. There are thrust efficiency losses with long tail pipes and structural difficulties with designing spars around the engines' bulk.  Boeing put the engines outside the wing because it's more efficient and moved them forward on pylons because they act as mass balances for the wing. Exhaust farther forward or farther aft doesn't make much difference. You are assuming the jet blast under the wing must create strong suction that inverts the lift force, but that's not how the lift is being made.

No, because the air is still turning the same way. It might even help a little bit, depending on the wing shape.

To create lift you need lower pressure above the wing than below, but the difference in speed is the effect, not cause of this. The cause is that the air would like to continue moving straight due to inertia, and the pressure decreases above until it can pull the air down along the surface, while it increases below until it can push the air out of its way.
The air behind the engine does not have lower pressure. Remember that the Bernoulli's equation is just formulation of conservation of energy for fluids. So it applies when the flow is accelerated without adding energy. Then the pressure decreases to compensate. But adding energy is exactly what the engine does. So in this case the increased kinetic energy of the flow is added by the engine, and the pressure does not decrease (well, it might, depending on the nozzle design and operating conditions, but the goal is that it does not, because that way the engine is most efficient).

To create a lift, the airflow speed on top of the wings should be higher than the airflow speed on the bottom of the wings.

That is correct. Lift is the result of a pressure difference between upper and lower side, and pressure is proportional to the inverse of speed squared if no energy is added.

But when you keep the engine on the bottom of the wings, wouldn't it hurt the lift as air flows faster on the bottom of the wings?

Note the point about adding energy in the paragraph above. As @JanHudec correctly points out, speed in the engine exhaust stream does not indicate suction. The pressure in the exhaust stream is still higher than on the upper side of the wing and lift is not diminished.
But friction drag is proportional to speed, so the higher speed on the upper side will cause more friction drag from the engine mount and nacelle. Placing the engine on the lower side puts it into a comparatively slow airstream. Also, the engine intake will slow down the air ahead of it (ram effect) and since the air that flows around the intake ends up on the lower side, this blocking effect of the engine intake causes less de- and acceleration of air, lowering losses. Adding the greatly increased volume stream of the hot exhaust gasses also keeps pressure up since all the volume behind the nacelle can be filled with exhaust gas. Adding the engine, therefore, slightly increases lift.
But the most important reason is maintenance and accessibility. The low-hanging engine is easy to reach and to inspect. That is a major reason why the "classic" layout that started with the Me-262 and was carried over to the passenger jets is still preferred. There have been designs with their engines mounted above the wing (for noise abatement), but that never caught on.
The next reason is cabin noise. By shielding the loud exhaust stream from the cabin by placing the wing between both, passenger comfort is significantly improved. This was especially important for the early jets with their high exhaust speeds.


text:
Was my latest Air Berlin flight in violation of some in-flight procedures?
I was quite surprised about one (all?) of doors on my latest Air Berlin flight. In particular:

the "Door locking indicator" was empty (black),
the "Cabin pressurized" light was not lighten and
the "Remove before flight" red strip was not removed.

(I am the owner of this picture and I authorize anyone to use it for any purpose except illegal activity.)
The plane landed safely and there was no noticeable in-flight change of air flow/pressure etc. Still, I wonder if the described situation is a violation of some in-flight procedures or at least odd in any way?

I would have asked a cabin attendant about it!

How long a flight was this? In other words, was it just an up and right down flight to get between two relatively close airports?

The door may have been inoperative, hence all of those things.  So long as the crew is aware of the situation, no violations would have occurred.

@mins: Sorry, I don't have. But, if I'm not mistaken then lever was as close to pin and red strip itself as possible.

@Terry. Nope. It was 6,5 hours long TXL->AUH flight. Picture was captured around the middle of this flight.

@Moo: If by "inoperative" you mean that these doors weren't use for boarding and leaving the plane then you're correct. These were close to the left wing doors.

@trejder I mean the door may have been marked on the maintenance list as "not to be used" - maybe a defective slide etc.  How full was the flight?

@Moo: I'm unable to access such information. Defective or not, shouldn't this pin be removed and "cabin pressurized" light turned on? This was an A330 flight nearly entirely full. In my (tail) section (where these doors were) I managed to see 2-3 empty chairs.

I did a Google search using "How many aircraft emergency exits can be inoperative?" and it appears that it's legal to dispatch some aircraft with one inoperative. For example, http://fsims.faa.gov/wdocs/policy%20letters/pl-001_r03_d00.htm. It appears that some restrictions may be required.

@Terry Assuming that doors being inoperative is the real cause and explanation for things observed by me -- this could be a good answer to my question. I'm still confused, however", why "cabin pressurized" light was not on. I assumed that it should be on whether cabin is pressurized (mine obviously was) irrespective of doors being inoperative or not.

@trejder are you talking about the "cabin pressurised" light on the door?  That may simply be part of the reason the door is inoperative...  As for removing the pin, not really as that is an indicator that the door is inoperative and will restrict the use of the door.

@Moo. OK, thanks. I think, that you / Terry / someone may gather together all the comments and formulate some kind of summary answer here.

I can't tell from this picture where the pin (and attached "remove before flight" strip) was actually placed. The thing does need to be kept somewhere during flight; it's not removed from the aircraft entirely.

@ZachLipton Yes. But, it shouldn't be kept inside the door, right?

It is usually attached to a different part of the door it belongs to.

The safety pin is in its stowage position - ie. where it is supposed to be stored when not in use, as you can see in this video. The red streamer is usually tucked away out of sight, but some airlines seem to leave it visible. Also, it appears from your photo that the arming handle is over to the right, ie. in the 'armed' position, as it should be in flight.
The Door Locking Indication is green-lettering-on-black for LOCKED, as you can see at 0:28 in this video, or red-lettering-on-black for UNLOCKED. But that lettering may not be very evident unless you are quite close to the door and looking almost directly at the indicator.
As far as I am aware, the Cabin Pressurized light is not normally illuminated in flight - it flashes red if-and-only-if the door handle is moved towards the OPEN position AND the cabin is pressurized, as you can see at 1:04 in this video.

You have managed to explain almost anything, thank you. As for the second part -- since I was staying in a quite long toilet queue I had pretty much enough time to actually stare directly into the "Door Locking Indicator" and I'm 100% positive that there was absolutely nothing inside, neither red nor green and no text. It was just a black, empty strip. Since we managed to figure out that these doors were inoperative, this may be a repercussion of that fact.


text:
Why does an engine bleed system failure restrict operating elevation?
Why are airplanes not permitted to operate to/from airports at over 9,200 ft if Engine Bleed Air Supply System #2 is inoperative?

That is not a limitation for all airplanes. What particular aircraft IS that a limitation for? Without this context, the question as posted makes no sense & will be closed as "Too Broad" or "Unclear What You're Asking."

The question is very unfortunately worded. There is a huge difference between limit to operating altitude, as most seem to have understood it, and limit to landing at or departing from airports at high altitude as the body of the question actually asks.

Without more details on the specific aircraft you are talking about, it isn't possible to give a very specific answer.
In general, if a commercial aircraft is unable to pressurize the cabin fully (usually to <8,000 ft, thanks @jwzumwalt, for both safety and comfort reasons), the operating ceiling is reduced to compensate. The human body doesn't play nicely with high altitudes, so to make sure that passengers and crew don't asphyxiate, the cabin must be pressurized to a safe level.
If one of the engine bleed air systems is inoperative, which among many other things provides for cabin pressurization, the remaining compressed air capacity is diverted to more mission-critical systems on the aircraft. This leaves the cabin pressurization system starved of the compressed air it needs, resulting in it not being able to fully pressurize the cabin to a safe and comfortable level.

FYI - In the US, cabin pressure must be at or below 8000ft for flights over 30min or oxygen must be made available to passengers. (14 CFR PART 121 SUPPLEMENTAL OXYGEN REQUIREMENTS.)

@jwzumwalt Thanks, I didn't know that. Learn something new every day!

The question is not asking about operating ceiling, but about operating to/from airports higher than 9,200 ft. Such airports are notable by their pattern altitude being higher than the standard limit for unpressurised flight at 10,000 ft (but note that nepal has at least two such airports serviced by unpressurised planes).


text:
Which books should I study to understand more about Supercritical airfoils?
I want to understand the theory behind supercritical airfoils and all the details about them. It would be nice if I could get a standard supercritical airfoil, its experimental Cl, Cd data in the transonic region and its verification from theory. I haven't been able to find anything concrete on the internet.

I don't know your answer, but I'd suggest looking at universities with classes on this topic, seeing if there are public copies of class syllabi on professors' or students' pages, and look for texts on those reading lists or course guides.

Good question and good answer. Unfortunately, if this question were asked today, I'm afraid it would be closed without a forethought...

Introduction to Transonic Aerodynamics
Roelof Vos, Saeed Farokhi (more)

Written to teach students the nature of transonic flow and its
  mathematical foundation, this book offers a much-needed introduction
  to transonic aerodynamics. The authors present a quantitative and
  qualitative assessment of subsonic, supersonic and transonic flow
  around bodies in two and three dimensions. The book reviews the
  governing equations and explores their applications and limitations as
  employed in modeling and computational fluid dynamics.

Transonic Aerodynamics of Airfoils and Wings
W.H. Mason (more)

Professor emeritus by the Virginia Tech, expert in aerospace
  engineering who worked at Grumman Aerospace where he served in the
  structural mechanics section and later in aerodynamics and worked on a
  number of high-profile projects, including the Grumman X-29. Has
  taught classes — including his "famous" Configuration Aerodynamics
  course.

Review of NASA Supercritical Airfoils
Richard T. Whitcomb (more)

NASA supercritical airfoils are characterized by a substantially
  reduced curvature of the midcord region of the upper surface together
  with increased camber near the trailing edge. The basic aerodynamic
  phenomena associated with the airfoils and representative wind tunnel
  results are discussed. The results indicate that the drag rise Mach
  numbers for NASA supercritical airfoils are 0.1 higher than for
  comparable NACA 6-series airfoils. A recent analytic method for
  predicting the aerodynamic characteristics of supercritical airfoils
  is described. The flight demonstration programs of three applications
  of supercritical airfoils utilizing the F-8, T-2C and F-111 as test
  beds are summarized.

Richard T. Whitcomb on Wikipedia.

NASA Supercritical Airfoils, A Matrix of Family-Related Airfoils
Charles D. Harris (more)

The NASA supercritical airfoil development program is summarized in a
  chronological fashion. Some of the airfoil design guidelines are
  discussed, and coordinates of a matrix of family related supercritical
  airfoils ranging from thicknesses of 2 to 18 percent and over a design
  lift coefficient range from 0 to 1.0 are presented.

Supercritical Airfoil Design for Future HALE Concepts
Kasim Biber, Carl P. Tilmann (more)

This paper presents design and analysis of a new airfoil for future
  High-Altitude Long-Endurance (HALE) aircraft that has an operational
  condition at supercritical speeds. The XFOIL and MSES computational
  codes were used to design, modify and analyze the airfoil.

Design of Laminar Supercritical Airfoils Based on Navier-Stokes Equations
Zhong-Hua HAN, Jie DENG, Jun LIU,  Ke-Shi, ZHANG, Wen-Ping SONG (more)

This paper aims to develop a suit of design methods and
  tools for laminar supercritical airfoils, based on
  high-fidelity computational fluid dynamics (CFD). The reliable
  transition perdition method, efficient inverse design and
  global optimization design methods are regarded as
  three key issues, which are addressed in this paper. The automatic
  transition prediction method coupling Navier-Stokes equations
  flow solver and eN linear
  stability analysis method is applied to the design and analysis of the
  laminar supercritical airfoils; inverse design and optimization design
  methods based on kriging surrogate model is developed to
  attain favorable pressure gradient while reducing wave
  drag; efficient optimization based on kriging model and an
  improved EI (expected improvement) method is developed for finding
  the global optimum.

NASA Technical Memorandum 103933 An Experimental Investigation of a Supercritical Airfoil at Transonic Speeds
George G. Mateer, H. Lee Seegmiller, Lawrence A. Hand and Joachim Szodruch (more)

Benchmark experimental data obtained in the two-dimensional, transonic
  flow field surrounding a supercritical airfoil are presented. Airfoil
  surface and tunnel wall pressure and LDV measurements are used to
  describe the flow on the model, above the wing and in the wake.
  Comparisons are made with calculations using the Reynolds-averaged
  Navier-Stokes equations. The results illustrate the performance of two
  turbulence models in both separated and attached flows. The largest
  differences between theory and experiment occurred in separated flows
  with the Johnson and King turbulence model providing the best
  estimates.

Analysis of Aerodynamic Characteristics of a Supercritical Airfoil for Low Speed Aircraft
P.Sethunathan, M.Niventhran, V.Siva, R.Sadhan Kumar (more)

The main goal of the proposed paper is the analysis of aerodynamic
  characteristics of an various supercritical airfoils like 0406, 0412,
  0706 and 1006 influence on the  dramatic improvement in lift and
  reduction of drag in low speed aircraft. In a precedent work,  a  cusp
  like  structure  at  the  trailing  edge  of  an  unsymmetrical
  airfoil  produces a  very  high  improvement  in  climbing performance
  of an aircraft.

Two-dimensional Wind-tunnel Tests of a NASA Supercritical Airfoil with Various High-lift Systems
E. Omar et Al. (more)

Three high lift systems for a 9.3 percent blunt based, supercritical
  airfoil were designed, fabricated, and wind tunnel tested. A method
  for calculating the viscous flow about two dimensional multicomponent
  airfoils was evaluated by comparing its predictions with test data. A
  comparison of high lift systems derived from supercritical airfoils
  with high lift systems derived from conventional airfoils is
  presented. The high lift systems for the supercritical airfoil were
  designed to achieve maximum lift and consisted of: (1) a single
  slotted flap, (2) a double slotted flap and a leading edge slat, and
  (3) a triple slotted flap and a leading edge slat.

Aerodynamic Characteristics of an 11-percent-thick Symmetrical Supercritical Airfoil at Mach Numbers Between 0.30 and 0.85
Blackwell, J. A., Jr. (more)

Abstract: The aerodynamic characteristics of an 11 percent thick
  symmetrical supercritical airfoil were determined in an 8 foot
  transonic pressure tunnel over a Mach number range of 0.30 to 0.85.
  The Reynolds number of the tests, based on the airfoil chord, varied
  with Mach number over a range of 360 million to 774 million. The
  geometric angle of attack varied from -0.5 to 10.5 deg. The abrupt
  drag rise for the supercritical airfoil at zero normal force
  conditions occurs at a Mach number just above 0.82. The corresponding
  drag rise Mach number for a conventional NACA 0012 airfoil is
  approximately 0.70. At zero normal force conditions, the level of
  supervelocity over the supercritical airfoil is considerably reduced
  from that for the NACA 0012 airfoil. Also, the shock wave for the
  supercritical airfoil is substantially weaker than that for the NACA
  0012 airfoil.

Hi mins, great answer. +10 if I could!

Really helpful answer!!

@mins can you tell me how do we know the design condition at which design lift is given which is used in the naming of the SC airfoils? Eg. SC(2) 0412.

@Manish: SC = supercritical, 2 = a familly in NASA-TP-2969 document, 04 = CL .4, 12 = 12% thickness.

Yes. And what are the conditions at which Cl = 0.4, Re, M, Angle of attack, are these specified for the given Cl?

@Manish: Ah ok, in general I don't know. For this particular one, see page 53 of this thesis.

Thanks a lot!! Just 1 last thing, how were you able to get hold of all these links and documents so efficiently? I tried and tried to search over net, but wasn't much successful. :)

@Manish: Many SC airfoils were developed in project named ATAT (Advanced Technology Airfoil Tests). You may search their server with ATAT and the airfoil as keyword. e.g. for SC(2) 0012 -- or -- SC(3) 0712. I've no recipe, just started  searching supercritical transonic Whitcomb and added more keywords. Google in that case helped by using my previous searches (usually this 'feature' sucks).

@mins why is Cl redesigned 0.5, not same as original Cl 0.4. Also, the coordinates available on the net are for original airfoil or redesigned?


text:
Do ducted fans eliminate induced drag?
It has often been said on this site that induced drag on wings is the inevitable result of producing lift over a finite length. Converting this to the realm of propulsion, I assume that propellers have "induced drag" or something similar, as they produce thrust over a finite length.
With a ducted fan, the airstreams above and below the airfoil can't just "slip around the side," because they're blocked by the duct. My intuition thus suggests that ducted fans would not have induced drag (or its equivalent), assuming that the space between the blades and the duct is negligible.
Is this accurate?

You need to read this answer to another question, especially the part on "Induced Drag" toward the end: http://aviation.stackexchange.com/a/13978/520

You are absolutely right, a propeller suffers from induced drag just as a wing does, and there are methods to minimize this induced drag. Simply put, you should distribute thrust elliptically over the prop blade's span and accelerate all air flowing through the prop disc by an equal amount. This was first found by A. Betz in 1919 and later refined by Larabee.
However, adding a duct will not change the basic reason why induced drag occurs. We had this discussion already here and here. A duct helps to equalize and decelerate the flow and is used in fan engines, because they prefer to operate in air flowing at a speed around Mach 0.5. But it will not help to reduce induced drag.

EDIT: I'll add this ahead of the answer because it will necessarily come up: This answer is based on the model described by Prandtl, which is only that: A model, a simplified description for what happens that works to explain some phenomena. There is some disagreement about this model, so a separate explanation using a differen theory might help. The problem is that an explanation with pressure differences doesn't help much here, at least I wouldn't know how to answer the question without working with Helmholtz/Prandtl.
Fundamentally, what we are talking about are vortex rings that are created because air is moved in some direction somewhere and not moved everywhere else. That's independent of how you move it, and fundamentally, a duct or winglet or something similar can't change that. You can look at the vortex itself as the thing that actually creates lift or thrust, so you want and need that vortex.
Then there's Helmholtz's theorems. Quick quote from that article:

Helmholtz’s first theorem:
The strength of a vortex filament is constant along its length.
Helmholtz’s second theorem:
A vortex filament cannot end in a fluid; it must extend to the
boundaries of the fluid or form a closed path.

In the case of a normal wing, the vortex ring ("closed path") consists of one side along the length of the wing, two sides trailing behind the wingtips and the fourth side is somewhat lost, or far enought away that it can mostly be ignored. The three sides that are not along the wing itself are the problem, they are what the term "induced drag" refers to. Because there is no surface at the end of the wing for the vortex to end at, it has to continue into infinity. And there is some friction involved in creating long fast-spinning vortices, so you get drag. (Any help in explaining this part more eloquently is appreciated)
Because the amount of lift/thrust that you want is given, the vortex strength isn't something that can be changed either. What can be changed, however, is the distribution/speed/radius. If the lift distribution is elliptical (see: Spitfire), the lift gradient is minimal across the wing so there is no strong and fast vortex at the end, it is gradually created along the whole wingspan and the induced drag is minimal. Winglets do something similar by forcing the vortex to go all the way around the winglet.
In the case of a propeller, the inner side of the vortex ends at the spinner or shaft, so the ring is broken and you are only left with a linear vortex, but it is still trailing infinitely into the air at the tips of the blades. You can look at a wing in the same way if you interrupt the ring at the fuselage.
In the case of a ducted fan, you have two fluid boundaries, both on the inside and on the outside, which appears to solve the problem. You can have a vortex around the blade only and no vortices trailing into infinity creating drag. The problem is that the fluid doesn't really end at the duct because the duct has a finite length. So what you get is a large vortex around the whole duct. Air is moving in at the front and out at the back, and somewhere, that air moves back to the front far away from the engine.
In the case of a ducted fan, the vortex ring is created in the place where the moving air meets the static air at the beginning and end of the duct. Instead of a very small vortex around the tip of the wing (and along the wing, to not mess up Helmholtz's theorems), you get a larger one around the whole duct.
Therefore, I wouldn't say that the duct can eliminate induced drag. The effect is fundamentally comparable to a wingtip device: It increases the inner radius of the vortex and thereby reduces the induced drag significantly. It's far more effective at that than a wingtip device, but it has to be because ducted fans are usually relatively small so the drag would otherwise be enormous.

Friction is what dissipates the vortex. The energy for creating it simply goes to kinetic energy of the air that is now moving in circles (drag is simply loosing energy).

In fact, the energy argument allows deriving the efficiency without considering vortices at all. Simply take the amount of air accelerated (down in case of wing, aft in case of propeller or fan) and it's speed and that gives you the impulse and energy per unit of time and those two give you the lifting/propulsion efficiency.

The energy argument allows you to not consider the vortices at all, but that's (in my opinion) exactly the problem with it here: To understand the effects of vortices, an approach that removes the vortices from the picture doesn't help. They are still there, they are just hidden inside of the control volume.

What's the vortex ring you keep talking about? What shape is it/where on the wing is it?

@JanHudec I'd really appreciate an energy/impulse argument. I've tried doing it out myself, but I've never been able to get something I'm satisfied with.

@raptortech97: Here is an illustration of what I'm talking about: https://tangvald.files.wordpress.com/2013/11/56.jpg

Thanks for the illustrations. What do you mean by an elliptical lift distribution?


text:
can the wreckages of a hot air balloon and a hang glider be salvaged for a working hang glider?
I'm currently working through my collection of Spike and Suzy albums, and in the album I'm currently reading, one of the main characters crashes with a hot air balloon near the wreckage of a hang glider without sail cloth. The main character, who is usually characterized as a somewhat smart man with skill in engineering, tore a large section from the balloon cloth, tied it to the hang glider wreckage and managed to make a functional hang glider that could even make a reasonable ascent in the Alps, near the Mont Blanc and catch up with his friends who were in other hot air balloons.
The hang glider looked like this (I couldn't find a bigger image, sorry):

How would this type of makeshift hang glider work IRL? even though hang gliders and hot air balloons are made from the same kind of fabric (ripstop nylon), wouldn't the holes between the frame and the cloth totally ruin the thing?

I'd at least fold the leading edge over the tube, the trailing edge isn't that important to be hole free

I know we have some glider folks around who might be able to answer this - at a quick glance I think that hang glider would be a bit tough to fly (for the reasons @ratchetfreak mentioned), but on a theoretical level it's sound. Some of the early flexible wing hang gliders aren't much more than what's illustrated there.

I'd also be a bit worried with bare hands at 15,780 feet, and temperatures as low as -40° F/C.

Has the hang glider actually been wrecked, or has the "sail" (cloth) simply rotted away or been removed?  If the former, the tubing surely has several kinks and dents that render it completely compromised strength-wise, and therefore un-airworthy even for low ground-skimming flights-- (ps the illustrated shape of airframe (metal parts) has as almost nothing in common with shape of actual hang glider airframe!)

To expand a bit on what @ratchetfreak said in a comment...
The leading edge on that glider is not secured properly.  It would need to go around the bar all the way along the leading edge of the wing, otherwise the holes would make it so the airfoil doesn't function.
As it is right now, the air would hit those gaps and make the cloth flutter, which would break up the airflow over the glider itself.  Thus you'd just have turbulence over the wing, which would make the wing stall.
So, yeah, in a nutshell, that wouldn't work as a glider as pictured above.
But, if he had found some string or wire from the wrecked balloon and properly secured some of the Hot Air Balloon cloth to the frame...  There's no reason it shouldn't work.  Hot Air Balloons are made out of fabric that is very very similar to what hang gliders are made from.